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A parallel plate capacitor with air betw...

A parallel plate capacitor with air between the plates has capacitance C. What will be the capacitance if
(a) the distance between the plates is doubled?
(b) the space between the plates is filled with a substance of dielectric constant 5?

Text Solution

Verified by Experts

Capacitance: `C=(Kepsi_(0)A)/(d)impliesCprop(1)/(d)` and `Cpro K`
(a)C/2 OR capacitance is halved .
(b)5C OR capacotance becomes 5 times the initial values.
Detailed Answer:
`C=(kepsi_(0)A)/(d)`
d=distance between the plates
A=area of the plates
K=dielectric constant
(a)`C=(epsi_(0)A)/(d)` (k=1 for air)
If the distance between the plates is doubled,then
`C.=(epsi_(0)A)/(2d)=(1)/(2)C`
So ,the capacitance will be halved.
(b)If dielectric of dielectric constant 5 is inserted then
`C.5=xx(epsi_(0)A)/(d)=5C`
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